Prove that $R[x_1,\ldots,x_n] / (x-r_1,\ldots,x-r_n) \cong R$

Question

I am reading Aluffi's "Algebra: Notes from the Underground" book and one of the exercises in the books asks the following:

Let $R$ be a commutative ring, and let $r_1, \ldots , r_n$ be elements of $R$.
Prove that $R[x_1,\ldots,x_n] / (x-r_1,\ldots,x-r_n) \cong R$.

I have tried to find an explicit isomorphism between the two sets in a similar fashion as to how one shows that $R[x] / (x-r) \cong R$, namely by using the function $e_r : R[x] \rightarrow R, f(x) \mapsto f(r)$, the First Isomorphism Theorem and the fact that $\ker(e_r)=(x-r)$ but I haven't been able to find a multivariable equivalent to $e_r$.

I have also tried the following: I know that $R[x_1,\ldots,x_n] \cong R[x_1,\ldots,x_{n-1}][x_n]$ and also that $R[x_1,\ldots,x_{n-1}][x_n] / (x-r_n) \cong R[x_1,\ldots,x_{n-1}]$ so $R[x_1,\ldots,x_n] / (x-r_n) \cong R[x_1,\ldots,x_{n-1}]$. Similarly, $R[x_1,\ldots,x_{n-1}] \cong R[x_1,\ldots,x_{n-2}][x_{n-1}]$ and $R[x_1,\ldots,x_{n-2}][x_{n-1}] /(x-r_{n-1}) \cong R[x_1,\ldots,x_{n-2}]$ so $$\bigl(R[x_1,\ldots,x_n]/ (x-r_n)\bigr)/(x-r_{n-1}) \cong R[x_1,\ldots,x_{n-2}] $$ and here I get stuck again, as I don't know how to make sense of the left-hand side of the last isomorphism.

So, I would appreciate if someone would tell me how to reason about this problem. Thanks.

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Note that this book adopts a rings-first approach and the tools available to the reader at this point reach the First and Third Isomorphism Theorem level.

Answer 1

Your approach is correct. For ideals $I,J \subseteq R$ of a commutative ring $R$ it holds $R/(I + J) \cong (R/I)/((I+J)/I)$. For example for $n = 2$: $$R[x_1,x_2]/(x_1-r_1,x_2-r_2) = R[x_1,x_2]/((x_1-r_1)+(x_2-r_2))\\ \cong (R[x_1,x_2]/(x_2-r_2))/((x_1-r_1)/((x_1-r_1)+(x_2-r_2))) \\ \cong R[x_1]/(x_1-r_1) \cong R.$$ The general case follows similarly by induction.

An explicit isomorphism is given by $R[x_1,\dots,x_n]/(x_1-r_1,\dots,x_n-r_n) \to R, \ f \mapsto f(r_1,\dots,r_n)$.

Answer 2

Here is a different approach, that avoids induction. There is a natural surjective homomorphism $\varphi:R[x_1,...,x_n]\to R$ given by $f\to f(r)$ where $r=(r_1,...,r_n)$. Let $I=(x_1-r_1,...,x_n-r_n)$. We just have to show that $I=\ker(\varphi)$. Clearly, $I$ is contained in the kernel. As for the reverse inclusion, there is a nice trick. Let $f\in\ker(\varphi)$, i.e $f(r)=0$. Write explicitly $f=\sum a_{i_1,...,i_n}x_1^{i_1}...x_n^{i_n}$. Now, by the definition of $I$, in the quotient ring $R[x_1,...,x_n]/I$ we have $x_i+I=r_i+I$ for each $i$. Hence:

$f+I=\sum a_{i_1,...,i_n}x_1^{i_1}...x_n^{i_n}+I=\sum a_{i_1,...,i_n}r_1^{i_1}...r_n^{i_n}+I=f(r)+I=0+I=I$

Which means $f\in I$, as required.